# Matrix Multiplication - Product of [Row or Column Vector] and Matrix [Lay P94, Strang P59]

From P59 of Intro to Lin Alg, 4th Ed by Strang & P94-95 of Linear Algebra and its Apps by Lay

For relief, I denote all row vectors with superscripts and column with subscripts. Define $\mathbf{A} = \left[\begin{matrix} \vec{a^1} \\ \vdots \\ \vec{a^i} \\ \vdots \\ \vec{a^m} \end{matrix}\right]_{m \times n} \& \quad \mathbf{B} = \left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p}$ ,
where each of the $m$ $\vec{a^i}$s has size $(1 \times n)$ and each of the $p$ $\vec{b_i}$s size $(n \times 1)$. Then:

$\mathbf{AB} = \mathbf{A}\left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p} = \left[\mathbf{A}\vec{b_1} \cdots \mathbf{A}\vec{b_i} \cdots \mathbf{A}\vec{b_p}\right]_{m \times p}. \tag{Row}$

$\text{ Also, } \qquad \mathbf{AB} = \left[\begin{matrix} \vec{a^1} \\ \vdots \\ \vec{a^i} \\ \vdots \\ \vec{a^m} \end{matrix}\right]\mathbf{B} = \left[\begin{matrix} \vec{a^1}\mathbf{B} \\ \vdots \\ \vec{a^i}\mathbf{B} \\ \vdots \\ \vec{a^m}\mathbf{B} \end{matrix}\right]_{m \times p}. \tag{Coln}$

$1.$ In (Row), how and why can $\mathbf{A}$ left-multiply into the column vector form of $\mathbf{B}$ ?
In (Coln), how and why can $\mathbf{B}$ right-multiply the row vector form of $\mathbf{A}$ ?

$2.$ Would someone please explain how $\mathbf{A}$ can be rewritten as a row vector?

• The expression $\mathbf{A} = \left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_n}\right]^{\huge{T}}_{m \times n}$ needs to be edited, as it states that $A$ is the transpose of a row vector. Commented Sep 2, 2013 at 5:16
• @littleO: Thanks; $\require{enclose} \enclose{horizontalstrike}{\vec{A_n}}$ should be $\vec{A_m}$. I wrote $\mathbf{A}$ with transpose to shuffle out of (gratuitous) elongation and white space. Is this not true: $\mathbf{A} = \left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]^{\huge{T}}_{m \times n} = \left[\begin{matrix} \vec{A_1} \\ \vdots \\ \vec{A_i} \\ \vdots \\ \vec{A_m} \end{matrix}\right]_{m \times n} ?$
– user53259
Commented Sep 2, 2013 at 8:51
• Actually $\left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]^{\huge{T}} = \left[\begin{matrix} \vec{A_1}^T \\ \vdots \\ \vec{A_i}^T \\ \vdots \\ \vec{A_m}^T \end{matrix}\right]$. Note that, if each $\vec{A_i}$ is $1 \times n$, then $\left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]$ is a row vector of size $1 \times mn$. Commented Sep 2, 2013 at 9:03
• I also adverted to math.caltech.edu/classes/ma1b-Pr/09Ma1bPracMatrixmult2.pdf
– user53259
Commented Dec 7, 2013 at 8:19

These things are all matrices, so it's standard matrix multiplication. For example,

$$\left[ \begin{matrix} 1 & 2\\ 3 & 4\\ \end{matrix}\right]\cdot \left[ \begin{matrix} 1 & 2\\ 3 & 4\\ \end{matrix}\right] = \left[ \begin{matrix} \left[ \begin{matrix} 1 & 2\\ 3 & 4\\ \end{matrix}\right]\cdot\left[ \begin{matrix} 1\\ 3\\ \end{matrix}\right] & \left[ \begin{matrix} 1 & 2\\ 3 & 4\\ \end{matrix}\right]\cdot\left[ \begin{matrix} 2\\ 4\\ \end{matrix}\right]\\ \end{matrix}\right] = \left[ \begin{matrix} \left[ \begin{matrix} 7\\ 15\\ \end{matrix}\right] & \left[ \begin{matrix} 10\\ 22\\ \end{matrix}\right]\\ \end{matrix}\right] =\left[ \begin{matrix} 7 & 10\\ 15 & 22\\ \end{matrix}\right].$$

(Of course, here, there is a slight notational abuse; the last two matrices aren't exactly the same).

• I think there is no abuse of notation, and the last two matrices are exactly the same. You're using block notation. Commented Sep 2, 2013 at 5:12
• @littleO One is a 1x2 matrix, while the other is a 2x2 matrix. Commented Sep 2, 2013 at 5:14
• That's not how I understand block notation. If $a_1,a_2$ are column vectors (let's say they're $N \times 1$), then I think $\begin{bmatrix} a_1 & a_2 \end{bmatrix}$ is just a concise notation for the $N \times 2$ matrix whose first column is $a_1$ and whose second column is $a_2$. In any case, I suppose the issue is minor. Commented Sep 2, 2013 at 5:20

The verification is trivial. But to give an intuitive explanation, remember this:

Left multiplication of $B$ by matrix $A$ is equivalent to perform a row transformation on $B$. Thus we have to guarantee $B$ contains enough rows to be transformed. Consider matrix is actually multilinear, to arrive at your equation $(Row)$.

With the similar reason, right-multiplication of $B$ by matrix $B$ is equivalent to performing a column transformation on $A$. Thus we must guarantee $A$ contains enough columns to be transformed. This effects your equation $(Column)$.

To see this, take a set of orthonormal basis $e_i,\ldots,e_n$. Remark the following facts.

• $Be_i=\begin{pmatrix}b(1, i)\\\vdots\\b_(n,i)\end{pmatrix}$ selects the $i$th column of matrix $B$
• $e_j^TA=\begin{bmatrix} A(j,1) & \cdots & A(j,n) \\ \end{bmatrix}$ selects the $j$th row of matrix $A$

Next write $\text{$i$th column of AB } = (AB)e_i=A(Be_i)=Ab_i$.
This means $i$th column of $AB$ is obtained from left-multiplying $A$ by the $i$th column of $B$.
So all the other columns ($\neq i$) of $B$ and $AB$ are unaffected.
Or we could say the left multiplication (by $\mathbf{A}$) is independent of the other columns of $B$.
Hence $A$ is actually a row transformation.

Similarly $\text{$j$th row of AB } = e_j^TAB = (e_j^TA)B = \begin{bmatrix} A(j,1) & \cdots & A(j,n) \\ \end{bmatrix}B = (\mathbf{A^j})^TB$.

• Thank you very much. +1. I wish I could've upvoted more. Could you please reveal how and why "left multiplication is independent of columns of $B$ and hence $A$ is actually a row transformation"? Also, how can I intuit or naturalise this? At present, I'm enduring/"capitulating" to this via the definition, but I'd like to discern it instinctively.
– user53259
Commented Dec 11, 2013 at 13:45
• @LePressentiment "the ith column of AB is obtained from left multiplicating A to the ith column of B", which is irrelvent to column $j\neq i$. That's what I mean "independent" Commented Dec 11, 2013 at 13:58
• I'd be grateful if you would please answer question $2$ also?
– user53259
Commented Mar 10, 2014 at 6:17
• @LePressentiment If $A$ is written as a row vector, then you cannot multiply $B$ anymore. Commented Mar 10, 2014 at 13:28